[LeetCode] 277. Find the Celebrity

Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the other n - 1 people know him/her but he/she does not know any of them.
Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B. You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
You are given a helper function bool knows(a, b) which tells you whether A knows B. Implement a function int findCelebrity(n), your function should minimize the number of calls to knows.
Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return -1.

Thought process:
Two pass. First pass find out the celebrity candidate. Second pass verify if candidate is truly a celebrity.
  1. First pass: set candidate = 0. For each person on the right, check if they know the candidate.
    1. If the person knows the candidate: continue.
    2. Otherwise, we know the candidate cannot be the celebrity. We also know the people before the current person cannot be the celebrity because they know the candidate. So update the candidate to be the current person.
    3. After the first pass, we know that only the candidate can be the celebrity, because the people before him/her cannot, and the people after can't be either because they know the candidate.
  2. Second pass: check if the candidate knows any of the other people.

Solution:
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/* The knows API is defined in the parent class Relation.
      boolean knows(int a, int b); */

public class Solution extends Relation {
    public int findCelebrity(int n) {
        int candidate = 0;
        for (int i = 1; i < n; i++) {
            if (!knows(i, candidate)) {
                candidate = i;
            }
        }
        
        for (int i = 0; i < n; i++) {
            if ((i != candidate && knows(candidate, i)) || !knows(i, candidate)) {
                return -1;
            }
        }
        return candidate;
    }
}

Time complexity: O(n) calls to knows.
Location: San Jose, CA, USA

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